MATH40003 Revised Historical Coverage Practice Paper 3

Question 1 Historical basis: 2023 Q1 and 2024 Q1. Directly restores ring axioms, explicit counterexamples, and matrix centres that were diluted in the old mocks. 30 minutes. Recognition signal: new operations or a request for the centre; return to each part of the definition. First key step: For a centre, commute a general element with matrix units. For “not a ring ” , it is enough to exhibit one failed axiom with explicit elements. Ordinary matrix addition is an abelian group; multiplication is associative; both distribu- tive laws follow entrywise. Hence Mn(F ) is a ring. In (b)(i), the operation ⊕ has no additive identity for negative integers: |−2+x| cannot equal −2. In (b)(ii), associativity fails. With A = diag(1, 2) and B = I, one has det ((det A)B) = 4 but (det A)(det B) = 2 , so (A ⊗ B) ⊗ C ̸= A ⊗ (B ⊗ C) for non-zero C. Write X = aE12 + bE13 + cE23 ∈ N3. Commuting with E12 gives c = 0, and commuting with E23 gives a = 0. The vector bE13 commutes with every element, so Z(N3) = span(E13). For X = (xij) ∈ Z(Mn(F )), commuting with every Ekk forces all off-diagonal entries of X to vanish. Thus X = diag(λ1, . . . , λn). Commuting with Eij gives λi = λj for all i, j, so X = λIn. Scalar matrices clearly commute with everything. Marking points: 4 ring; 4 counterexamples; 6 centre of N3; 6 centre of full matrix ring. Common errors: Checking only commutativity of addition; giving a map that is not even a binary operation; proving only that scalar matrices are central, not the converse. Final self-check: For the centre calculations, multiply with E12 and E23 explicitly; for Mn, verify the scalar result works in both directions. 2/5 MATH40003 - Linear Algebra and Group Theory Revised Historical Coverage

Question 2 Historical basis: 2023 Q3. The official-answer codomain error in the old analysis is avoided: every constructed image lies in U ∩ W . 30 minutes. Recognition signal: two three-dimensional subspaces share two basis vectors, and one must construct or rule out linear maps with specified kernel/image properties. First key step: Use uniqueness of coordinates in the basis (b1, b2, u, w). Linear maps are then defined by choosing images of these four basis vectors. If x ∈ U ∩ W , write x = αb1 + βb2 + γu = α′b1 + β′b2 + δw. Uniqueness of basis coordinates gives γ = δ = 0, so U ∩W = span(b1, b2). Also U +W contains all four basis vectors, hence equals V , and 3 + 3 = 2 + 4 verifies the dimension formula. For (c), define f (b1) = b1, f (b2) = b2, f (u) = f (w) = 0 . Then ker f = span(u, w), so nullity is 2, and f (U ∩ W ) = U ∩ W . If g : V → V is surjective, rank-nullity makes it injective. Since 0 ̸= b1 ∈ U ∩ W , the vector g(b1) is non-zero and lies in g(U ) ∩ g(W ), contradiction. For (e), define h(b1) = h(b2) = 0 , h (u) = b1, h (w) = b2. Then im h = span(b1, b2), so rank is 2, while h(U ) = span(b1) and h(W ) = span(b2) have zero intersection. Marking points: 4 intersection; 3 sum/dimension; 4 first construction; 4 impossibility; 5 second construction. Common errors: Defining an image outside the stated codomain; checking only spanning or only independence; forgetting that surjective endomorphisms of a finite-dimensional space are injective. Final self-check: Write each map as a 2 × 4 coordinate matrix and verify rank/nullity directly . 3/5 MATH40003 - Linear Algebra and Group Theory Revised Historical Coverage

Question 3 Historical basis: 2024 Q3. Course method is direct bilinear expansion and standard matrix represen- tation, not exterior algebra. 30 minutes. Recognition signal: “bilinear + beta(v ,v)=0”indicates expansion of β(v+w, v+w) and a skew-symmetric matrix. First key step: Expand the alternating condition first; then use β(v, w) = vT Aw in a basis. Since 0 = β(v + w, v + w), 0 = β(v, v) + β(v, w) + β(w, v) + β(w, w) = β(v, w) + β(w, v), so the form is skew-symmetric. With matrix A, alternation gives vT Av = 0 for all v. Taking v = ei gives aii = 0 , and taking v = ei + ej gives aij + aji = 0. Hence AT = −A. The converse follows because the scalar vT Av equals its transpose vT AT v = −vT Av, and over R this implies it is zero. In R3 the matrices are   0 a b −a 0 c −b −c 0   . Their determinant is zero. If the matrix is non-zero, one of a, b, c is non-zero and the corre- sponding 2 × 2 principal minor has determinant a2, b2 or c2, so the rank is exactly 2. In dimension two the same argument gives A = λJ. The matrix J has determinant 1, and direct multiplication gives vT J v = 0. Marking points: 4 skew result; 5 matrix test; 5 rank in R3; 6 classification in R2. Common errors: Using 2−1 without stating the field; saying an odd skew matrix has zero deter- minant without explaining why or giving the rank lower bound; confusing symmetric and alternating. Final self-check: Check AT = −A entry by entry and calculate the displayed 2 × 2 minors. 4/5 MATH40003 - Linear Algebra and Group Theory Revised Historical Coverage

Question 4 Historical basis: 2023 Q5-Q6 and 2024 Q5-Q6. It combines dihedral relations, automorphism groups, inner automorphisms and commutators, but each part uses the official direct group method. 30 minutes. Recognition signal: rks, conjugation maps, Aut(G), and the commutator subgroup. First key step: Establish sr = r−1s first. It reduces every product/inverse in D to one of two normal forms. The relation srs = r−1 follows by checking the action on 1, . . . ,6. Hence srk = r−ks. The powers rk are distinct. Also no rk can equal rℓs: otherwise s = rk−ℓ, but s fixes the vertex 1, whereas every non-identity power of the 6-cycle r has no fixed point, and s ̸= e. Finally rks = rℓs implies rk = rℓ, hence k = ℓ. Using srk = r−ks, products and inverses of the two normal forms stay in D, so the subgroup test applies. The identity automorphism is in Aut (G), composition is associative, the composition of isomorphisms is an isomorphism, and the inverse of an isomorphism is a homomorphism; hence Aut (G) is a group. Each Cg is an automorphism with inverse Cg−1, and Cgh = Cg ◦ Ch. Moreover Cg is the identity iff gx = xg for every x, so the kernel is Z(G). Conjugation sends a commutator to a commutator: g[x, y]g−1 = [gxg −1, gyg −1]. Therefore it sends every product of commutators into the same subgroup, so [G, G] ⊴ G. Finally xy[G, G] = yx[G, G] because (yx)−1xy = x−1y−1xy is a commutator . Thus the quotient is abelian. Marking points: 7 dihedral subgroup; 3 Aut group; 5 inner homomorphism/kernel; 3 normality; 2 abelian quotient. Common errors: Assuming r and s commute; checking closure only for one of the four product types; proving properties only for a single commutator rather than products of them. Final self-check: Reduce srk to r−ks; check Cgh(x) on an arbitrary x; verify the quotient equality with the coset criterion. 5/5